Algebra · Algebraic proof
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Proving divisibility results
The expand–collect–factorise engine: to prove "always a multiple of k", exhibit the expression as k × (whole number) — and say so.
Algebra · Algebraic proof
Proving divisibility results
The expand–collect–factorise engine: to prove "always a multiple of k", exhibit the expression as k × (whole number) — and say so.
Why it works
"A multiple of " MEANS " times a whole number". So a divisibility proof has one goal: wrestle the expression into the exact shapeand then say that's what it is. The engine is always the same: encode → expand → collect → factorise out → conclude.
Watch it run — *the sum of five consecutive integers is a multiple of 5*:
is a whole number, so the sum is 5 × (whole number) — a multiple of 5, for every . The factorised line is the proof; the sentence after it is the mark scheme's final tick. Stopping at leaves the reader to do the last step — write the form explicitly.
Sometimes everything cancels — proving a constant. Some "show that" claims reduce to a number: expand carefully and watch the 's annihilate. If the survivors are just , you have proven the expression ALWAYS equals 2 — independence from is the point, and sign errors in the expansion are the only enemy.
Disproving divisibility: exhibit the leftover. "The sum of four consecutive integers is a multiple of 4"? Encode and expand: — there is always a remainder of 2, so it is NEVER a multiple of 4. (A single numeric counterexample, , also suffices — but the algebra shows it fails for every , which is stronger.)
The grade-9 tool: products of consecutive integers. Among any two consecutive integers one is even; among any three, one is a multiple of 3 (and one is even). So is always even, and is always divisible by both 2 and 3 — hence by 6. Factorising INTO consecutive pieces, then citing these facts, cracks problems that pure expansion can't.